Elementary Geometry for College Students (7th Edition) Clone

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 5 - Section 5.4 - The Pythagorean Theorem - Exercises - Page 253: 32

Answer

$AD=13$

Work Step by Step

In $\triangle ABC$ $AC^{2}=AB^{2}+BC^{2}$ $AC^{2}=4^{2}+3^{2}$ so AC = 5 In $\triangle ACD$ $AD^{2}=AC^{2}+DC^{2}$ $AD^{2}=5^{2}+12^{2}$ $AD^{2}=25+144$ $AD^{2}=169$ $AD=13$
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